Question

1. A file of size 500 MB is to be transferred from host H1 to host H2 through routers R1, R2 and
R3. See Figure 1 below.

h1=2mbs=>R1==8mbs=>R2==10mbs==>R3=1mbs==>h2
(a) If the file is transmitted is one entity, without dividing it into smaller chunks, estimate
how long it would take for the file to arrive at the receiver. [4 marks]
(b) If the file is first divided into smaller chunks, and each chunk is transmitted and then
combined, estimate how long it would take for the whole file to arrive at the receiver.
(Hint: if a file is divided into smaller chunks, transfer time is estimated based on the
bottleneck link).

Answer 1

a)

Consider givend data:R1 = 500 kbps, R2 = 2 Mbps, and R3 = 1 Mbps

The throughput for the file transfer=min{R1,R2,R3}

=min{500 kbps, 2 Mbps, 1 Mbps}

   =500 kbps

So, the throughput for the file transfer=500 kbps

b)

Consider given data:

The file size= 4 million bytes

Convert million bytes to bits

=32000000 bits.

From (a), Throughput for the file transfer=500 Kbps

=500000 bps

Dividing the file size by the throughput,roughly how long will it take to transfer the file to Host B:

=file size/hroughput for the file transfer

=32000000 bits/500000 bps

=64 seconds

c)

Consider the given data:

Repeat (a) and (b), but now with R2 reduced to 100 kbps.

That means R2=100 kbps , R1 = 500 kbps, and R3 = 1 Mbps

The throughput for the file transfer=min{R1,R2,R3}

=min{500 kbps, 100 kbps, 1 Mbps}

   =100 kbps

So, the throughput for the file transfer=100 kbps

Dividing the file size by the throughput,roughly how long will it take to transfer the file to Host B:

=file size/hroughput for the file transfer

=32000000 bits/100000 bps

=320 seconds